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3 years ago

What is ( √2 - √3 )² =

Mathematics

2 answers:

madreJ [45]3 years ago

7 0

2-2(square root of 6)+ 3
=5-2(square root of 6)
or 0.1

allochka39001 [22]3 years ago

4 0

just use FOIL (First, Outer, Inner, Last).

( √2 - √3 ) X ( √2 - √3 )

First: √2 X √2 = 2
Outer: √2 X -√3 = -√6
Inner: -√3 X √2 = -√6
Last: -√3 X -√3 = 3

and then add them together.

2 - √6 - √6 + 3
5 - 2(√6)

The answer is 5 - 2(√6) .

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State if the two triangles are congruent. If they are, state how you know.

Gwar [14]

Answer:

The triangles can be congruent.

Step-by-step explanation:

They are congruent if proven by SSS: 2 sides are clearly stated that they are congruent due to the marks they have.

The last side can be congruent if the diagonals are congruent in length by proving.

They can also be congeuent due to SAS because there is gonna be alternate interior angles due to the transversal.

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3 years ago

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3 years ago

Construct 3 linear equation starting with qiven solution z = 1/3

Andreyy89

Answer:

(a) $9z+2=5$

(b) $21z-11=-4$

Step-by-step explanation:

We are required to construct 3 linear equations starting with the given solution z = 1/3.

Equation 1

$z=\frac{1}{3}$

Multiply both sides by 9

$9z=\frac{1}{3}\times 9\\9z=3$

Rewrite 3 as 5-2

9z=5-2

Add 2 to both sides

Our first equation is: $9z+2=5$

Equation 2

$z=\frac{1}{3}$

Multiply both sides by 21

$21z=\frac{1}{3}\times 21\\21z=7$

Rewrite 7 as 11-4

21z=11-4

Subtract 11 from both sides

Our second equation is: $21z-11=-4$

Equation 3

$z=\frac{1}{3}$

Multiply both sides by 6

$6z=\frac{1}{3}\times 6\\6z=2$

Rewrite 6z as 4z+2z

4z+2z=2

Subtract 2z from both sides

Our third equation is: $4z=2-2z$

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3 years ago

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jenyasd209 [6]

The second answer is x=1

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3 years ago

Find the exact value of cos(7\pi /12)

Naya [18.7K]

7π/12 lies in the second quadrant, so we expect cos(7π/12) to be negative.

Recall that

$\cos^2x=\dfrac{1+\cos(2x)}2$

which tells us

$\cos\left(\dfrac{7\pi}{12}\right)=-\sqrt{\dfrac{1+\cos\left(\frac{7\pi}6\right)}2}$

Now,

$\cos\left(\dfrac{7\pi}6\right)=-\cos\left(\dfrac\pi6\right)=-\dfrac{\sqrt3}2$

and so

$\cos\left(\dfrac{7\pi}{12}\right)=-\sqrt{\dfrac{1-\frac{\sqrt3}2}2}=\boxed{-\dfrac{\sqrt{2-\sqrt3}}2}$

7 0

3 years ago